The 2nd, 31st and last term of an A.P. is, \(\frac{1}{2}\) and − \(

1.	The 2nd, 31st and last term of an A.P. is, \(\frac{1}{2}\) and − \(6\frac{1}{2}\) respectively. The number of terms of the A.P. is(a) 48 (b) 60 (c) 52 (d) 59
Answer» Answer : (d) 59 Let the first term, common difference and number of terms of the A.P. be a, d and n respectively. Then, a + d = \(7\frac{3}{4}\) ... (i) a + 30d = \(\frac{1}{2}\) ...(ii) and a + (n – 1)d = − \(6\frac{1}{2}\) ...(iii) Eqn (ii) – Eqn (i) ⇒ 29d = \(\frac{1}{2}\) - \(\frac{31}{4}\) = - \(\frac{29}{4}\) ⇒ d = - \(\frac{1}{4}\) Putting d = - \(\frac{1}{4}\) in (i), we get a - \(\frac{1}{4}\) = \(7\frac{3}{4}\) + \(\frac{1}{4}\) ⇒ a = 8 ∴ Putting the values of a and d in (iii), we have 8 + (n – 1) ( - \(\frac{1}{4}\) ) = - \(\frac{13}{2}\) ⇒ 8 + \(\frac{1}{4}\)- \(\frac{1}{4}\)n = - \(\frac{13}{2}\) ⇒ −\(\frac{1}{4}\)n = − \(\frac{13}{2}\) − \(\frac{33}{4}\) ⇒ − \(\frac{1}{4}\)n = − \(\frac{59}{4}\) ⇒ n = 59

The 2nd, 31st and last term of an A.P. is, \(\frac{1}{2}\) and − \(6\frac{1}{2}\) respectively. The number of terms of the A.P. is(a) 48 (b) 60 (c) 52 (d) 59

Answer»

Answer : (d) 59

Let the first term, common difference and number of terms of the A.P. be a, d and n respectively.

Then,

a + d = \(7\frac{3}{4}\) ... (i)

a + 30d = \(\frac{1}{2}\) ...(ii)

and a + (n – 1)d = − \(6\frac{1}{2}\) ...(iii)

Eqn (ii) – Eqn (i)

⇒ 29d = \(\frac{1}{2}\) - \(\frac{31}{4}\) = - \(\frac{29}{4}\)

⇒ d = - \(\frac{1}{4}\)

Putting d = - \(\frac{1}{4}\) in (i), we get

a - \(\frac{1}{4}\) = \(7\frac{3}{4}\) + \(\frac{1}{4}\)

⇒ a = 8

∴ Putting the values of a and d in (iii), we have

8 + (n – 1) ( - \(\frac{1}{4}\) ) = - \(\frac{13}{2}\)

⇒ 8 + \(\frac{1}{4}\)- \(\frac{1}{4}\)n = - \(\frac{13}{2}\)

⇒ −\(\frac{1}{4}\)n = − \(\frac{13}{2}\) − \(\frac{33}{4}\)

⇒ − \(\frac{1}{4}\)n = − \(\frac{59}{4}\)

⇒ n = 59